# Cos x and Sin x

jackson_sun
I have been asked to differentiate cos x and six....the maclaurin series versions...

I have done the general and specific terms as shown below.

Im not sure if these are correct?
thanks

General Terms
cos x = ∑ (-1)n (x^(2n) / (2n)!)

COS x = ∑ (-1)n (x^(2n+1)/ (2n +1) / (2n)!)

sin x = ∑ (-1)n (x^(2n+1) / (2n+1)!)

‘sin x = ∑ (-1)n (((2n+1) x^(2n))/ (2n+1)!)

Specific Terms
sin x = x – (x^3 / 3!) + (x^5 / 5!) – (x^7 / 7!) + (x^9 / 9!) - …
‘sin x = 1 – (3x^2 / 3!) + (5x^4 / 5!) - (7x^6 / 7!) + (9x^8 / 9!) - …

cos x = 1 – (x^2 / 2!) + (x^4 / 4!) – (x^6 / 6!) + (x^8 / 8!) - …
COS x = x – (x^3 / 3 / 2!) + (x^5 / 5 / 4!) – (x^7 / 7 / 6!) + (x^9 / 9 / 8!) - …
thanks

Homework Helper
You already posted the same thing else where, learn what the factorial actually means! Then you can simplify yours derivatives and realize what your not realizing right now!

Homework Helper
I have been asked to differentiate cos x and six....the maclaurin series versions...

I have done the general and specific terms as shown below.

Im not sure if these are correct?
thanks

General Terms
cos x = ∑ (-1)n (x^(2n) / (2n)!)

COS x = ∑ (-1)n (x^(2n+1)/ (2n +1) / (2n)!)

You know to use "^" to indicate exponent of the "x"- use for the exponent on (-1) also: (-1)^n. Obviously "cos x" can't be equal to both of these. Do you mean cos' x? Also A/B/C, at best, ambiguous. Do you mean
((x^(2n+1)/(2n+1))/(2n)! or (x^(2n+1))/((2n+1)/(2n)!)?

Finally, if you are differentiating as you say, you are "going the wrong way"! The derivative of x^n is nx^(n-1), not (1/(n+1)) x^(n+1).

sin x = ∑ (-1)n (x^(2n+1) / (2n+1)!)

‘sin x = ∑ (-1)n (((2n+1) x^(2n))/ (2n+1)!)

Specific Terms
sin x = x – (x^3 / 3!) + (x^5 / 5!) – (x^7 / 7!) + (x^9 / 9!) - …
‘sin x = 1 – (3x^2 / 3!) + (5x^4 / 5!) - (7x^6 / 7!) + (9x^8 / 9!) - …

cos x = 1 – (x^2 / 2!) + (x^4 / 4!) – (x^6 / 6!) + (x^8 / 8!) - …
COS x = x – (x^3 / 3 / 2!) + (x^5 / 5 / 4!) – (x^7 / 7 / 6!) + (x^9 / 9 / 8!) - …
thanks

Homework Helper
It looks to me the cos x in capitals is the integral of cos x in normal...

jackson_sun
yeh COS x is the integral of cos x....i meant integral of cos x and derivative of sin x...sorry for the confusion.

I have realised how to simplify 'sin x so that it equals cos x....but i am not sure how to simplify the integral of cos x to equal sin x.

COS x = x – (x^3 / 3 / 2!) + (x^5 / 5 / 4!) – (x^7 / 7 / 6!) + (x^9 / 9 / 8!)
= ?

any idea? and on the subject...what is the notation for the integral of cos x...I have used COS x but this is obviously incorrect.
thanks...

Homework Helper
Take your second term...if you don't see the arithmetic, don't try this level of math yet, $$\frac{\frac{x^3}{3}}{2!} =\frac{x^3}{3\cdot2!} = \frac{x^3}{3!}$$.

I didn't take into account the sign of the integral, but you get what i mean.

To learn how to type the math graphics im typing, click on the images I make appear. That will tell you what i typed to get that. Or at least say, "Integral of.."...

jackson_sun
I understand exactly what you have done.
How would it be done for the general terms is my bigger problem

Homework Helper
I understand exactly what you have done.

Good
How would it be done for the general terms is my bigger problem

$$\cos x = \sum_{n=0}^{\infty} -1^n \frac{x^{2n}}{2n!}$$

Integrate every term in the sum, with respect to x. In every individual term, n is constant.

So

$$\int\cos x dx= \sum_{n=0}^{\infty} -1^n \frac{\frac{x^{2n+1}}{2n+1}}{2n!}$$

(2n+1)2n!=(2n+1)!,

therefore it simplifies to

$$\int\cos x dx= \sum_{n=0}^{\infty} -1^n \frac{x^{2n+1}}{(2n+1)!}$$, which is exactly your series for sin x.

Homework Helper
Just one comment: it would be far better to write
$$\int\cos x dx= \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!}$$
with parentheses around the -1. I would be inclined to interpret -1n as -(1n) which, of course, would be just -1!

Homework Helper
O yes i guess i would too, if someone else wrote that..thanks

jackson_sun
Thanks gents

Homework Helper
No Problem, and welcome to PF